On 5/6/12 at 8:28 PM, jack.j.jepper at googlemail.com (J.Jack.J.) wrote:
>Thanks for replying. Responses embedded:
>On May 6, 8:24 am, Murray Eisenberg <mur... at math.umass.edu> wrote:
>>First, perhaps folks were reluctant to respond because this looked
>>like it could be a homework exercise.
>>Second, you don't even have proper Mathematica syntax in your
>>equation relating x and k. Did you even try to read the
>>documentation to learn the very basics?
>I tried and tried for hours but couldn't so much as find any section
>that would even tell me how to write the condition that k be an
>integer.
The specifics of how to specify k is an integer depends on
exactly what function you will be using. For several functions
such as Integrate, you can use Assuming to specify assumptions
about variables. For example, compare:
In[12]:= Assuming[{m, n} \[Element] Integers && m != n,
Integrate[Cos[n x] Cos[m x], {x, \[Pi], 0}]]
Out[12]= 0
with the result obtained with no assumptions about m,n
In[13]:= Integrate[Cos[n x] Cos[m x], {x, \[Pi], 0}]
Out[13]= (n cos(\[Pi] m) sin(\[Pi] n)-m sin(\[Pi] m) cos(\[Pi] n))/(m^2-n^2)
In other cases such as for NMinimize or NMaximize, conditions
for variables are specified as part of the syntax. Details can
be found in the documentation.
But, specifying k is an integer in your particular case really
isn't useful.
>For example, proper syntax for the equation would be:
>>(x/Log[x]) (1 + 1/Log[x]) == 108.2 + k
>>Third, the equation itself looks really nasty. Aside from the fact
>>that it mixes exact formulas with an approximate real (108.2), the
>>left-hand side is transcendental.
This is a key point. Typically, transcendental equations do not
have closed form symbolic solutions and are only solved
numerically. In general, when there is a closed form symbolic
solution, it is because the equation occurs often enough in some
field someone created a special function to represent the solution.
So, assuming the only possible solution is a numerical solution
you will have to substitute specific values for k to get a
solution such as was done with:
>>FindRoot[Evaluate[f[x] /. k -> 2], {x, 1.1}] {x -> 1.11137}